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Java Stack error after getting last element via pop method

I have the following code for the solution of Hackerrank.
public static void main(String[] args) {
System.out.println(isBalanced("{(([])[])[]}"));
}
public static String isBalanced(String s) {
Stack<Character> stack = new Stack<>();
stack.push(s.charAt(0));
for (int i = 1; i < s.length(); i++) {
Character c = s.charAt(i);
Character cStack = stack.peek();
if (cStack == '{' && c == '}'
|| cStack == '[' && c == ']'
|| cStack == '(' && c == ')') {
stack.pop();
} else {
stack.push(c);
}
}
if (stack.isEmpty())
return "YES";
return "NO";
}
Although the code seems to working without any problem, it throws the following error on the Hackerrank page. I already test the input in my local IDE as it is {(([])[])[]}, but I am not sure if I need to get the last element (it maybe due to getting it via Character cStack = stack.peek(); and then stack.pop();.
So, could you please have a look at and test this code on Hackerrank page and let me know what is wrong?
Update:
public static String isBalanced(String s) {
Stack<Character> stack = new Stack<>();
stack.push(s.charAt(0));
for (int i = 1; i < s.length(); i++) {
Character c = s.charAt(i);
if (c == '{' || c == '[' || c == '(') {
stack.push(c);
} else if (stack != null) {
Character cStack = stack.peek();
if (cStack == '{' && c == '}'
|| cStack == '[' && c == ']'
|| cStack == '(' && c == ')') {
stack.pop();
}
}
}
if (stack.isEmpty())
return "YES";
return "NO";
}

Before calling stack.peek(), you need to check if the stack is empty or not. Calling pop() or peek() on an empty stack will raise an error.
If the current character is an opening bracket, you don't even need to check the stack top. If it is a closing bracket, then check if the stack is empty or not first. If it is, return false. Otherwise compare the top character and make a decision.

Palindrome Stack

I wrote this method to check to see if a words is a palindrome, but when it is a palindrome it keeps returning false
public boolean isPalindrome(String s){
int i;
int n = s.length();
Stack <Character> stack = new Stack <Character>();
for (i = 0; i < n/2; i++)
stack.push(s.charAt(i));
if (n%2 == 1)
i++;
while(!stack.empty( )) {
char c = stack.pop( );
if (c != s.charAt(i));
return false;
}
return true;
}

I'm not sure why you're not using { } brackets. Try to learn proper Java conventions early.
if (c != s.charAt(i)); // <- this semicolon is your problem
return false;
Is equivalent to:
if (c != s.charAt(i)) {
// Do nothing
}
// Do this no matter what
return false;
Furthermore, the logic on your for-loop may be flawed for similar reasons. Remove your semicolon, and better yet, practice always using brackets:
if (c != s.charAt(i)) {
return false;
}
#jhamon also points out that you never actually increment i in your while loop:
while(!stack.empty( )) {
char c = stack.pop( );
if (c != s.charAt(i)) {
return false;
}
i++;
}

Trying to implement a stack and the bracket matching problem but cannot figure out some semantic errors

I am required to write my own stack using an array which then I have to implement the bracket matching problem.
This is my algorithm: if we're given the string: "(())" it is returning "not balanced" when it should say it is "balanced". If you look over at BracketCheck.java near the end it tests whether the stack is empty where if it is it should return true and say the given string: "(())" is balanced. I am having problems debugging it and the result of the operation is incorrect.
Here is my stack and main class code:
public class Stack {
// instance fields:
private char array[]; // our array
private int topOfStack; // this indicates the value for each position
// overloaded constructor:
public Stack(int size) {
this.array = new char[size]; // here we instantiate a new char type array.
this.topOfStack = -1; // we set the starting position of the topOfStack to -1.
}
// push method:
public void push(char character) {
if(isFull() == true) {
System.out.println("Stack overflow error.");
}
else {
topOfStack++;
array[topOfStack] = character;
}
}
// pop method:
public char pop() {
if(isEmpty() == true) {
//System.out.println("Overflow.");
return '\0';
}
else {
char c = array[topOfStack];
topOfStack--;
return c;
}
}
// top method:
public int top() {
if(isEmpty() == true) {
System.out.println("The stack is empty.");
return 0;
}
else {
return array[topOfStack]; // returns the top element in the stack.
}
}
// size method:
public int size() {
return topOfStack + 1;
}
// isEmpty method:
public boolean isEmpty() {
if(topOfStack == -1) {
return true;
}
else {
return false;
}
}
// isFull method:
public boolean isFull() {
if(topOfStack == array.length -1 ) {
System.out.println("Stack if full.");
return true;
}
else {
return false;
}
}
BracketCheck.java
public class BracketCheck {
public static void main(String args[]) {
String text = "(())";
//boolean check = isBalanced(text);
if(isBalanced(text) == true) {
System.out.println("Balanced.");
}
else {
System.out.println("Not balanced");
}
}
public static boolean isBalanced(String text) {
Stack stack = new Stack(25); // for simplicity our stack is going to be a size of 25.
char[]arr = text.toCharArray(); // convert our string to a set of characters in an array.
// here we are looping through our text
for(int i = 0; i < arr.length; i++) {
if(arr[i] == '{' || arr[i] == '(' || arr[i] == '[') {
stack.push(arr[i]);
}
else if(arr[i] == '}' || arr[i] == ')' || arr[i] == ']') {
if(stack.isEmpty()) {
return false; // if empty then return false.
}
else if((stack.pop() == '(' && arr[i] != ')') || (stack.pop() == '{' && arr[i] != '}') || (stack.pop() == '[' && arr[i] != ']')) {
return false; // here we return false because this tells us that there is a mismatch and not balanced. Also if we did pop out the brace and it was equal
// to it's corresponding closing brace then it will just continue the loop other return false to indicate it is unbalanced.
}
}
}
if(stack.isEmpty()) {
return true; // after looping through the text if the stack turns out to be empty then this shows that there the text is balanced indeed because
}
else {
return false;
}
}
}

In this condition
else if((stack.pop() == '(' && arr[i] != ')') || (stack.pop() == '{' && arr[i] != '}') || (stack.pop() == '[' && arr[i] != ']')) {
You pop 3 times, so you extract 3 different elements from your stack. To fix it, pop once and save popped element in local variable:
if(stack.isEmpty()) {
return false; // if empty then return false.
}
else {
char element = stack.pop();
if((element == '(' && arr[i] != ')') || (element == '{' && arr[i] != '}') || (element == '[' && arr[i] != ']')) {
return false;
}
}

Analyze a string of characters in java for format A^nB^n

I have a string of characters with As and Bs that I need to analyze for a Language A^nB^n. I can use the following code to work most of the time but when there is a letter that is not an "A" or "B" it may still return true, for example: AABACABAA should not be true, but it says it is. AABB is true; AABBAABB is not true. I have to use stacks and am not allowed to use counting.
public static boolean isL2(String line){
// set up empty stacks
Stack L2Stack = new Stack();
// initialize loop counter
int i = 0;
int n = line.length();
/* Push all 'A's to a_stack */
while ((i < line.length()) && (line.charAt(i) == 'A')) {
char ch = line.charAt(i);
L2Stack.push(ch);
i++;
}
/* Pop an 'A' for each consecutive 'B' */
while ((i < line.length()) && (line.charAt(i) == 'B')) {
if (!L2Stack.empty()){
L2Stack.pop();
i++;
}
else
return false;
}
if (i == n && !L2Stack.empty()){
return false; // more As than Bs
}
if (i != n && L2Stack.empty()){
return false; //more Bs than As
}else
return true;
}

if (i != n && L2Stack.empty()) {
return false; //more Bs than As
}
Should be
if (i != n) {
return false;
}
Since if you haven't finished reading all the characters, you can't return true, regardless of whether or not the stack is empty.
I'm assuming that AAABBBA should return false.
That change would also handle illegal characters.

Testing parentheses in a equation using stack java

I'm writing a program that will take in an equation and check if all the parentheses line up and it will output if it is good or not.
For Ex: (3+4) is good
((3*8) is NOT Good
I'm not allowed to use java's built in push() pop() methods ext..
I have to make my own which I think I got....I think!
The problem I'm having is in the Test() method.
First I'm not sure how to write the while loop like:
while(there are still characters)
Anyway the output I'm getting is: stack is empty -1
Any help is appreciated. I'm one of the slower program learners and I couldn't be trying any harder. Thanks.
Here's what I got:
public class Stacked {
int top;
char stack[];
int maxLen;
public Stacked(int max) {
top = -1;
maxLen = max;
stack = new char[maxLen];
}
public void push(char item) {
top++;
stack[top] = item;
}
public int pop() {
//x = stack[top];
//top = top - 1;
top--;
return stack[top];
}
public boolean isStackEmpty() {
if(top == -1) {
System.out.println("Stack is empty" + top);
return true;
} else
return false;
}
public void reset() {
top = -1;
}
public void showStack() {
System.out.println(" ");
System.out.println("Stack Contents...");
for(int j = top; j > -1; j--){
System.out.println(stack[j]);
}
System.out.println(" ");
}
public void showStack0toTop() {
System.out.println(" ");
System.out.println("Stack Contents...");
for(int j=0; j>=top; j++){
System.out.println(stack[j]);
}
System.out.println(" ");
}
//}
public boolean test(String p ){
boolean balanced = false;
balanced = false;
//while ( )
for(char i = '('; i < p.length(); i++ ){
push('(');
}
for (char j = ')'; j < p.length(); j++){
pop();
}
if (isStackEmpty()) {
balanced = true;
//return balanced;
}
return balanced;
}
public static void main(String[] args) {
Stacked stacks = new Stacked(100);
String y = new String("(((1+2)*3)");
stacks.test(y);
//System.out.println(stacks.test(y));
}
}
Now I'm getting somewhere. I need to be pointed in the right direction again. Thanks everyone this helped big time. I still have a lot more to do but this is good for now. Eventually I need to create a two more methods: one "infix to postfix" and the other "evaluating postfix" and at the end I'll need to read in answers from a text file instead of putting my own into the main method. Thanks again much appreciated.

Unless you need to actually evaluate the equation, a stack is too complicated a solution here. You simply need a counter:
int openParentheses = 0;
for (int i = 0; i < p.length(); i++) {
if (p.charAt(i) == '(') {
openParentheses++;
} else if (p.charAt(i) == ')') {
openParentheses--;
}
//check if there are more closed than open
if (openParentheses < 0) {
return false;
}
}
if (openParentheses == 0) {
return true;
} else {
return false;
}
If you absolutely must use stacks, use this:
for (int i = 0; i < p.length(); i++) {
if (p.charAt(i) == '(') {
push('x'); //doesn't matter what character you push on to the stack
} else if (p.charAt(i) == ')') {
pop();
}
//check if there are more closed than open
if (stackIsEmpty()) {
return false;
}
}
if (isStackEmpty()) {
return true;
} else {
return false;
}

I agree with Griff except that you should include another check if you didn't have more closed parentheses than open. (x*y))( is not a valid entry.
int openParentheses = 0;
for (int i = 0; i < p.length(); i++) {
if (p.charAt(i) == '(') {
openParentheses++;
} else if (p.charAt(i) == ')') {
openParentheses--;
}
if(openParentheses<0)
return false;
}
if (openParentheses == 0) {
return true;
} else {
return false;
}

You may be required to use a stack, but this could be done with a simple counter. This will show you a how to iterate over the characters of a String:
boolean test(String p) {
int balance = 0;
for (int idx = 0; idx < p.length(); ++idx) {
char ch = p.charAt(idx);
if (ch == '(')
++balance;
else if (ch == ')')
--balance;
if (balance < 0)
return false;
}
return balance == 0;
}
Of course, you could replace the increment and decrement with pushes and pops, respectively, on a stack.

For parsing you can use a for loop over the index and address the character of the string at the certain index.
But you actually do not need a stack, an integer variable openBraces is sufficient:
initialize with 0
for '(' you increment the variable one
for ')' you decrement the variable one
if openBraces is <0, you immediately give an error
if at the end openBraces is not equal to 0, you give an error.
Since you should do your homework yourself, I did not post source code, only explanations ;)

I think you just need this --
for ( int i = 0 ; i < p.length(); i++ ) {
char c = p.charAt(i);
if ( c == '(' )
push('(');
else if ( c == ')' ) {
if ( isStackEmpty() ) {
// Return error here because of unbalanced close paranthesis
}
pop();
}
else {
// do nothing
}
}
You CAN use a stack if you must, but considering how simplistic this is, you just need a counter that you increment and decrement and check for 0 at the end.
If you do use a counter, you should check after every decrement if the value is less than 0. If so, throw an error.
Edited based on Ryan/Dave Ball's comments.

It could be done like this:
String equation = "(2+3))";
Integer counter = 0;
//while(equation)
for(int i=0; i<equation.length();i++)
{
if(equation.charAt(i)=='(')
{
counter++;
}
else
if(equation.charAt(i)==')')
{
counter--;
}
}
if(counter == 0)
{
System.out.println("Is good!!!");
}
else
{
System.out.println("Not good!!!");
}
}